ZOJ Problem Set - 3019 Puzzle

Puzzle

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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For sequences of integers a and b, if you can make the two sequences the same by deleting some elements in a and b, we call the remaining sequence "the common sub sequence". And we call the longest one the LCS.

Now you are given two sequences of integers a and b. You can arrange elements in a and b in any order. You are to calculate the max length of the LCS of each arrangement of a andb.

Input

Input will consist of multiple test cases. The first line of each case is two integers N(0 < N < 10000), M(0 < M < 10000) indicating the length of a and b. The second line is N 32-bit signed integers in a. The third line is M 32-bit signed integers in b.

Output

Each case one line. The max length of the LCS of each arrangement of a and b.

Sample Input

5 4
1 2 3 2 1
1 4 2 1

Sample Output

3

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Author: ZHUANG, Junyuan

Source: ZOJ Monthly, August 2008

分析：

开始看到LCS(最长公共子序列)还以为要用dp，仔细一看，原来You can arrange elements in a and b in any order.（a，b中元素的顺序可以任意排），这样问题就更简单了。直接sort后进行模拟。

ac代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
using namespace std;
#define M 10001
#define N 10001
 int a[M],b[N];
int main(){
    int x,y,i,j,n;
   while(cin>>x>>y)
   {
       for(i=0;i<x;i++){
        cin>>a[i];
    }
    for(j=0;j<y;j++){
        cin>>b[j];
    }
    sort(a,a+x);
    sort(b,b+y);
    n=0,i=j=0;
    while(i!=x&&j!=y){
        if(a[i]==b[j]){
            i++;
            j++;
            n++;
        }
        else if(a[i]<b[j]) i++;
        else j++;
    }
    cout<<n<<endl;
   }
    return 0;
}